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Consider again the system x _ = Ax = _1 1 1 3 _ x (36) that we discussed in Example 2

ISBN: 9781119256007 392

Solution for problem 15 Chapter 7.8

Elementary Differential Equations and Boundary Value Problems | 11th Edition

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Problem 15

Consider again the system x _ = Ax = _1 1 1 3 _ x (36) that we discussed in Example 2. We found there that A has a double eigenvalue r1 = r2 = 2 with a single independent eigenvector (1) = (1, 1) T , or any nonzero multiple thereof. Thus one solution of the system (36) is x(1) (t) = (1) e2t and a second independent solution has the form x(2) (t) = te2t + e2t , where and satisfy (A 2I) = 0, (A 2I) = . (37) In the text we solved the first equation for and then the second equation for . Here, we ask you to proceed in the reverse order. a. Show that satisfies (A 2I)2 = 0. b. Show that (A 2I)2 = 0. Thus the generalized eigenvector can be chosen arbitrarily, except that it must be independent of (1) . c. Let = (0, 1) T . Then determine from the second of equations (37) and observe that = (1, 1) T = (1) . This choice of reproduces the solution found in Example 2. d. Let = (1, 0) T and determine the corresponding eigenvector . e. Let = (k1, k2) T , where k1 and k2 are arbitrary numbers. What condition on k1 and k2 ensures that and (1) are linearly independent? Then determine . How is it related to the eigenvector (1) ? 1

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ISBN: 9781119256007

Since the solution to 15 from 7.8 chapter was answered, more than 218 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 75 chapters, and 1655 solutions. This textbook survival guide was created for the textbook: Elementary Differential Equations and Boundary Value Problems, edition: 11. The answer to “Consider again the system x _ = Ax = _1 1 1 3 _ x (36) that we discussed in Example 2. We found there that A has a double eigenvalue r1 = r2 = 2 with a single independent eigenvector (1) = (1, 1) T , or any nonzero multiple thereof. Thus one solution of the system (36) is x(1) (t) = (1) e2t and a second independent solution has the form x(2) (t) = te2t + e2t , where and satisfy (A 2I) = 0, (A 2I) = . (37) In the text we solved the first equation for and then the second equation for . Here, we ask you to proceed in the reverse order. a. Show that satisfies (A 2I)2 = 0. b. Show that (A 2I)2 = 0. Thus the generalized eigenvector can be chosen arbitrarily, except that it must be independent of (1) . c. Let = (0, 1) T . Then determine from the second of equations (37) and observe that = (1, 1) T = (1) . This choice of reproduces the solution found in Example 2. d. Let = (1, 0) T and determine the corresponding eigenvector . e. Let = (k1, k2) T , where k1 and k2 are arbitrary numbers. What condition on k1 and k2 ensures that and (1) are linearly independent? Then determine . How is it related to the eigenvector (1) ? 1” is broken down into a number of easy to follow steps, and 237 words. Elementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9781119256007. The full step-by-step solution to problem: 15 from chapter: 7.8 was answered by , our top Math solution expert on 03/13/18, 08:17PM.

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